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Lecture 4
1. Gauss’ Law
+ Examples
2. Conductors in Electric Field
Gauss’ Law
Electric Flux
We have used electric field lines to visualize electric fields and indicate their strength.
We are now going to count the number of electric field lines passing through a surface, and use this count to determine the electric field.
E
The electric flux passing through a surface is the number of electric field lines that pass through it.
Because electric field lines are drawn arbitrarily, we quantify electric flux like this: E=EA, except that…
If the surface is tilted, fewer lines cut the surface.
E A
Later we’ll learn about magnetic flux, which is why we will use the subscript E on electric flux.
E
E
A
The “amount of surface” perpendicular to the electric field is A cos .
A = A cos so E = EA = EA cos .
We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.
Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos .
Remember the dot product? E E A
If the electric field is not uniform, or the surface is not flat…
divide the surface into infinitesimally small surface elements and add the flux through each…
dA E
iE i i
A 0i
lim E A
E E dA
A
If the surface is closed (completely encloses a volume)…
E
…we count* lines going out as positive and lines going in as negative…
E E dA dA
a surface integral, therefore a double integral
Gauss’ Law
Mathematically*, we express the idea two slides back as
enclosedE
o
qE dA
Gauss’ Law
We will find that Gauss law gives a simple way to calculate electric fields for charge distributions that exhibit a high degree of symmetry…
… and save more complex and realistic charge distributions for advanced classes.
To see how this works, let’s do an example.
Example: use Gauss’ Law to calculate the electric field from an isolated point charge q.
To apply Gauss’ Law, we construct a “Gaussian Surface” enclosing the charge.
The Gaussian surface should mimic the symmetry of the charge distribution.
For this example, choose for our Gaussian surface a sphere of radius r, with the point charge at the center.
Let’s work the rest of the example ………
Strategy for Solving Gauss’ Law Problems
Select a Gaussian surface with symmetry that matches the charge distribution.
Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface.
Evaluate the surface integral (electric flux).
Determine the charge inside the Gaussian surface.
Solve for E.
Use symmetry to determine the direction of E on the Gaussian surface.
Worked Example 1
Compute the electric flux through a cylinder with an axis parallel to the electric
field direction.
E
The flux through the curved surface is zero since E is perpendicular
to dA there. For the ends, the surfaces are perpendicular to E, and E
and A are parallel. Thus the flux through the left end (into the
cylinder) is –EA, while the flux through right end (out of the cylinder)
is +EA. Hence the net flux through the cylinder is zero.
A
Gauss’s Law
Gauss’s Law relates the electric flux through a closed surface
with the charge Qin inside that surface.
0
inE
QE dA
This is a useful tool for simply determining the electric
field, but only for certain situations where the charge
distribution is either rather simple or possesses a high
degree of symmetry.
Worked Example 2
Starting with Gauss’s law, calculate the electric
field due to an isolated point charge q.
q E
r dA
We choose a Gaussian surface that is a
sphere of radius r centered on the point
charge. I have chosen the charge to be
positive so the field is radial outward by
symmetry and therefore everywhere
perpendicular to the Gaussian surface.
E dA E dA Gauss’s law then gives:
0 0
inQ qE dA E dA
Symmetry tells us that the field is
constant on the Gaussian surface.
2
2 2
0 0
14 so
4e
q q qE dA E dA E r E k
r r
Worked Example 3
An insulating sphere of radius a has a uniform charge density ρ and a total
positive charge Q. Calculate the electric field outside the sphere.
a
Since the charge distribution is spherically
symmetric we select a spherical Gaussian
surface of radius r > a centered on the
charged sphere. Since the charged sphere
has a positive charge, the field will be
directed radially outward. On the Gaussian
sphere E is always parallel to dA, and is
constant. Q
r E
dA
2Left side: 4E dA E dA E dA E r
0 0
Right side: inQ Q
2
2 2
0 0
14 or
4e
Q Q QE r E k
r r
Worked Example 3 cont’d
a
Q
Find the electric field at a point inside the sphere.
Now we select a spherical Gaussian surface
with radius r < a. Again the symmetry of the
charge distribution allows us to simply evaluate
the left side of Gauss’s law just as before. r
The charge inside the Gaussian sphere is no longer Q. If we
call the Gaussian sphere volume V’ then
2Left side: 4E dA E dA E dA E r
3
2
0 0
44
3
inQ rE r
34
Right side: 3
inQ V r
3
3 3230 00
4 1 but so
43 43 4
3
e
r Q Q QE r E r k r
a ar a
Worked Example 3 cont’d
2
3
We found for ,
and for ,
e
e
Qr a E k
r
k Qr a E r
a
a
Q
Let’s plot this: E
r a
Conductors in Electrostatic
Equilibrium
• The electric field is zero everywhere inside
the conductor
• Any net charge resides on the conductor’s
surface
• The electric field just outside a charged
conductor is perpendicular to the conductor’s
surface
By electrostatic equilibrium we mean a situation
where there is no net motion of charge within the
conductor
Conductors in Electrostatic
Equilibrium
Why is this so?
If there was a field in the conductor the charges
would accelerate under the action of the field.
• The electric field is zero everywhere inside the
conductor
++
++
++
++
++
++
---------------------
Ein
E E
The charges in the conductor
move creating an internal
electric field that cancels the
applied field on the inside of
the conductor
The charge on the right is twice the magnitude of the
charge on the left (and opposite in sign), so there are
twice as many field lines, and they point towards the
charge rather than away from it.
Combinations of charges. Note that, while the lines are less
dense where the field is weaker, the field is not necessarily
zero where there are no lines. In fact, there is only one point
within the figures below where the field is zero – can you
find it?
When electric charges are at rest, the electric
field within a conductor is zero.
The electric field is always perpendicular to the
surface of a conductor – if it weren’t, the
charges would move along the surface.
The electric field is stronger where the surface is
more sharply curved.
Worked Example 4 Any net charge on an isolated conductor must reside on its surface and
the electric field just outside a charged conductor is perpendicular to its
surface (and has magnitude σ/ε0). Use Gauss’s law to show this.
For an arbitrarily shaped conductor we
can draw a Gaussian surface inside the
conductor. Since we have shown that the
electric field inside an isolated conductor
is zero, the field at every point on the
Gaussian surface must be zero.
From Gauss’s law we then conclude that the
net charge inside the Gaussian surface is
zero. Since the surface can be made
arbitrarily close to the surface of the
conductor, any net charge must reside on the conductor’s surface.
0
inQE dA
Worked Example 4 cont’d
We can also use Gauss’s law to determine the electric field just outside
the surface of a charged conductor. Assume the surface charge density
is σ. Since the field inside the conductor is
zero there is no flux through the face
of the cylinder inside the conductor. If
E had a component along the surface
of the conductor then the free charges
would move under the action of the
field creating surface currents. Thus E
is perpendicular to the conductor’s
surface, and the flux through the
cylindrical surface must be zero.
Consequently the net flux through the
cylinder is EA and Gauss’s law gives:
0 0 0
orinE
Q AEA E
Worked Example 5
A charged rod of radius a has a net charge +Q per unit length
distributed uniformly over its surface. Find the Electric field at any point
outside the cylinder. Use Gauss’s law to find the electric field
everywhere, and to determine the charge distribution on the a shell.
a
b
Based on symmetry, E is everywhere directed radially outwards. If we
chose the Gaussian surface with a cylinder radius r, and of unit length h.
On the curved surface, E is constant magnitude and perpendicular to the
surface. The flux of E is therefore equal to the produce E times the
surface area. As h is unit length:
2r E = Q/o => E = Q/2r o
+Q
Maxwell’s first of four equations!!
Gauss' law can be written in a mathematically elegant form by formally representing the electric flux through a closed surface, S, as a surface integral.
Gauss’s law now gives the first of
Maxwell’s equations:
QdSES
r .0
S
QdSD.
But, we know that: D = E
r
E
0
. D.
This also gives rise to Laplace
Equation which states: r
V
0
2 .
Summary
• Two methods for calculating electric field – Coulomb’s Law
– Gauss’s Law
• Gauss’s Law: Easy, elegant method for symmetric charge distributions
• Coulomb’s Law: Other cases
• Gauss’s Law and Coulomb’s Law are equivalent for electric fields produced by static charges